$Assertion :$ $A$ bubble comes from the bottom of a lake to the top.
$Reason :$ Its radius increases.

When $n$ identical mercury drops combine to form a single big drop,

What is the excess pressure inside a bubble of soap solution of radius $5.00 \; mm$,given that the surface tension of soap solution at the temperature $(20 \; ^{\circ}C)$ is $2.50 \times 10^{-2} \; N m^{-1}$? If an air bubble of the same dimension were formed at a depth of $40.0 \; cm$ inside a container containing the soap solution (of relative density $1.20$),what would be the pressure inside the bubble? ($1$ atmospheric pressure is $1.01 \times 10^{5} \; Pa$).

The excess pressure inside a soap bubble of radius $0.5 \ cm$ is balanced by the pressure due to an oil column of height $4 \ mm$. If the density of the oil is $900 \ kg \ m^{-3}$, then the surface tension of the soap solution is (Acceleration due to gravity $= 10 \ m \ s^{-2}$)

An air bubble doubles its radius on rising from the bottom of a water reservoir to the surface. If the atmospheric pressure is equal to $10 \, m$ of water,the height of the water in the reservoir is ..... $m$.

$A$ large number of water droplets each of radius '$r$' combine to form a large drop of radius '$R$'. If the surface tension of water is '$T$' and the mechanical equivalent of heat is '$J$',then the rise in temperature due to this process is: